Ummm….yeah….that’s not the difference between Markovnikov and anti-Markovnikov addition.

Those of you who’ve taken sophomore organic chemistry will remember that Markovnikov’s rule states that protic functional groups (e.g., H-NR’R’) will typically add to double bonds such that the hydrogen adds to the less substituted carbon (and the other group, e.g., -NR’R’, adds to the more substituted carbon.) What Hartwig and coworkers have drawn as the Markovnikov product is still anti-Markovnikov. The products they’ve drawn might charitably be called conformational isomers, but they’ve failed to note any 3-D structure.

Of course, the true Markovnikov product would place the amine on the same carbon as the R group.

11 Responses to “WWWTP? – Markovnikov Edition”

This seems like something not very important. It’s a webpage hidden under a link which will not be visited by anybody except some specialists pretty much, and they are obviously going to know the difference between the two products. It’s a really horrible mistake in a textbook. But on a CENTC webpage, it’s a mistake you can safely ignore in order to get to what you wanted when you visited the page, which is the article list.

Full disclosure: I have a CENTC pen. I use it from time to time but it’s not a very reassuring pen because the metal part with the ball is too long and exposed, and I always feel like I’m on the point of breaking it.

I’m just going to put this out there. I really, really hate markovnikov’s rule.
Why? Because it encourages people to learn rules without thinking about the chemistry behind them.
I find it better to consider the underlying electronic argument. Electrophilic attack on an alkene always proceeds via the most stable cationic intermediate (or the transition state in which the developing charge is best stabilised)
Rather than, for example phrasing it as ‘the hydrogen adds to the end with the most hydrogens already’ it is better to say ‘the more negative (nucleophilic) atom adds to the carbon best able to stabilise positive charge’
For example, adding H-X to isobutene: in this case the hydrogen is essentially H+, and the X is adding as X-. If you protonate the alkene on the terminal CH2 , you get a stable tertiary cation intermediate to which the X- can add. However, if you added the H to the central carbon, you would get a drastically unstable primary cation. So far, so good for Markovnikov.
But what about hydroboration? This is classically ‘anti-markovnikov’, but why? In this case the H is essentially adding as H-, and the boron is more positive (electrophilic), so the theory is exactly the same.
The best thing about this argument is that it can be extended to non-hydrogen atoms, and it obviates the need to ‘remember’ which reactions go by ‘markovnikov’ mechanisms, and which are ‘anti-markovnikov’

Phillip, your analogy would mean that a terminal carbanion is formed after H- addition. It’s not a very good way to think of things. Anti-Markovnikov doesn’t happen in the normal mechanism and that’s it. Also a terminal carbocation can rearrange to form a more stable species as well…

Hydroboration is concerted and organometallic catalysis would work via a different route where terminal reductive elimination is favored due to sterics around the intermediate. It just doesn’t work the same way so you can’t use the same argument.

@eugene, I think I may have explained myself badly. Sorry. At no point did I mean to imply a full anion. I meant H- as a formalism to say it was acting as nucleophilic hydrogen rather than electrophilic

Perhaps hydroboration is a bad example – my meaning was that the build-up of positive charge in the transition state of the B-H addition is on the carbon best able to stabilise positive charge, as well as being sterically less hindered. In this case the steric and electronic arguments are reinforcing

I simply want to highlight the fact that adding H-X to an alkene is no different from any other electrophilic addition. Consider the reaction of a general molecule X-Y to an alkene. Which way round would it add? Markovnikov cannot help you as there is no ‘H’. Considering the electronics of the molecule and reaction – going by the most stable carbocation intermediate – provides the correct answer, and prevents students from thinking there is something ‘special’ about hydrogen or acids.